Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, Q;
int d[110000];
int parent[110000];
int q[110000];
vector<int> g[110000];
basic_string<int> mn[110000][20];
int ancestor[110000][20];
int LCA(int u, int v) {
  if (d[u] < d[v]) {
    swap(u, v);
  }
  for (int i = 19; i >= 0; i--) {
    if (d[u] - (1 << i) >= d[v]) {
      u = ancestor[u][i];
    }
  }
  if (u == v) {
    return u;
  }
  for (int i = 19; i >= 0; i--) {
    if (ancestor[u][i] != ancestor[v][i]) {
      u = ancestor[u][i];
      v = ancestor[v][i];
    }
  }
  return parent[u];
}
basic_string<int> query(int u, int v) {
  int o = LCA(u, v);
  basic_string<int> ret = mn[o][0];
  int du = d[u] - d[o];
  for (int i = 0; i < 20; i++) {
    if (du & 1 << i) {
      ret += mn[u][i];
      u = ancestor[u][i];
    }
  }
  int dv = d[v] - d[o];
  for (int i = 0; i < 20; i++) {
    if (dv & 1 << i) {
      ret += mn[v][i];
      v = ancestor[v][i];
    }
  }
  sort(ret.begin(), ret.end());
  return ret;
}
int main() {
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    scanf("%d", &c);
    mn[c][0].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(mn[i][0].begin(), mn[i][0].end());
    if (mn[i][0].size() > 10) {
      mn[i][0].resize(10);
    }
  }
  q[0] = 1;
  for (int i = 0, tail = 1; i < n; i++) {
    int u = q[i];
    for (int it = 0; it < g[u].size(); it++) {
      int v = g[u][it];
      if (v == parent[u]) {
        continue;
      }
      d[v] = d[u] + 1;
      parent[v] = u;
      q[tail++] = v;
    }
  }
  for (int i = 1; i <= n; i++) {
    ancestor[i][0] = parent[i];
  }
  for (int k = 1; k < 20; k++) {
    for (int i = 1; i <= n; i++) {
      ancestor[i][k] = ancestor[ancestor[i][k - 1]][k - 1];
      mn[i][k] = mn[i][k - 1] + mn[ancestor[i][k - 1]][k - 1];
      sort(mn[i][k].begin(), mn[i][k].end());
      if (mn[i][k].size() > 10) {
        mn[i][k].resize(10);
      }
    }
  }
  while (Q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    basic_string<int> res = query(u, v);
    if (res.size() > a) {
      res.resize(a);
    }
    printf("%d ", (int)res.size());
    for (int i = 0; i < res.size(); i++) {
      printf("%d ", res[i]);
    }
    printf("\n");
  }
  return 0;
}